Question 726562
{{{(-3x+5)/(15x^2+2x-1)}}}
The vertical asymptotes will occur for x values that make the denominator zero (and are not "holes"). To find them, factor the numerator and denominator:
{{{(1*(-3x+5))/((5x-1)(3x+1))}}}
If there was a common factor between the numerator and denominator, then there would be a "hole". We do not have a common factor so there are no holes. Now we can proceed to determine the vertical asymptotes. We set the denominator equal to zero:
(5x-1)(3x+1) = 0
and solve for x. From the Zero Product Property:
5x-1 = 0 or 3x+1 = 0
Solving we get:
x = 1/5 or x = -1/3
These are the equations for the vertical asymptotes.<br>
For the horizontal asymptotes, we start with the original function:
{{{(-3x+5)/(15x^2+2x-1)}}}
and divide each term by the highest power of x. This would be {{{x^2}}}:
{{{(-3/x+5/x^2)/(15+2/x-1/x^2)}}}
Now we look at this and figure out what happens when x becomes larger and larger (both positively and negatively). As x becomes large each of those fractions (which have an x in the denominator and a constant in the numerator) will become smaller and smaller. IOW: closer and closer to zero. So the numerator will become close to zero the the denominator will become close to 15. 0/15 is zero so the horizontal asymptote is y = 0.<br>
FWIW: Here's what the graph looks like:
{{{graph(500, 500, -5, 5, -10, 10, (-3x+5)/(15x^2+2x-1))}}}